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# Squeeze Theorem

Top
 Sub Topics Calculus is one of the substantial branches of mathematics deal with various important topics such as limits, integration, differentiation. Theorems based on calculus generate very important results help to solve various complicated problems in few steps. Squeeze theorem is one of them. The squeeze theorem is a theorem regarding the limit of a function. This theorem is also known as the sandwich theorem, the pinching theorem, the squeeze lemma, the sandwich rule or Kathy Theorem. Before start working with Squeeze theorem, let us know about limit of function, in calculus the limit of a function is a fundamental concept that depends upon the behavior of that function near a given or particular point. Limit of a function to tend to positive infinity, to minus infinity, or to a real limit, as x tends to positive infinity or to minus infinity also a function to tend to a real limit as x tends to a given real number. The squeeze theorem is frequently used in mathematical analysis and calculus. In this page will learn about this theorem in detail.

## Definition

The Squeeze theorem is employed on limit problems the place that the usual algebraic techniques certainly not successful. However, it requires that you manage to squeeze your problem in between two other simpler functions whose limits can be easily computable. In other words, the Squeeze theorem concerns the limit of a function that is squeezed between two other functions, each of which has the same limit at a given value.

If M(x) $\leq$ G(x) $\leq$ N(x) for all x in an open interval containing a, except possibly at a itself, and if

Lim$_{x \rightarrow a}$ M(x) = L = Lim$_{x \rightarrow a}$ N(x)
Then Lim$_{x \rightarrow a}$ G(x) exists and equal to L i.e.

Lim$_{x \rightarrow a}$ G (x) = L.

## Proof

To prove sequeeze theorem, consider the limit superior and inferior:

L = Lim$_{x \rightarrow a}$M(x) $\leq$ Lim$_{x \rightarrow a}$ Inf$\leq$ G(x) $\leq$ Lim$_{x \rightarrow a}$SupG(x) $\leq$ Lim$_{x \rightarrow a}$N(x) = L

Consider ($\epsilon$, $\delta$), for all $\epsilon$ > 0 there exists $\delta$ > 0 such that
0 < |x - a| < $\delta$ (for all x)

=> -$\epsilon$ < G(x) - L < $\epsilon$
As Lim$_{x \rightarrow a}$ M(x) = L,

for all $\epsilon$, $\epsilon$ > 0 there exists $\delta_1$ > 0
and for all x, 0 < |x - a| < $\delta_1$

=> -$\epsilon$ < M(x) - L < $\epsilon$  ...(1)

Similarly, Lim$_{x \rightarrow a}$ N(x) = L,
for all $\epsilon$, $\epsilon$ > 0 there exists $\delta_2$ > 0

and for all x, 0 < |x - a| < $\delta_2$
=> -$\epsilon$ < N(x) - L < $\epsilon$  ...(2)

M(x) $\leq$ G(x) $\leq$ N(x)
or M(x) - L $\leq$ G(x) - L $\leq$ N(x) - L

If $\delta$ = min($\delta_1$, $\delta_2$), then if |x - a| < $\delta$
From equation (1) and (2), we obtain

-$\epsilon$ < M(x) - L $\leq$ G(x) - L $\leq$ N(x) - L < $\epsilon$

or -$\epsilon$ < G(x) - L < $\epsilon$
which shows that  Lim$_{x \rightarrow a}$ G(x) = L.

## Examples

Few examples on the Squeeze theorem are given below:

Example 1: Prove that Lim$_{y \rightarrow 0}$ y sin ($\frac{1}{y}$) = 0 using squeeze theorem.

Solution: Let G(y) = y sin ($\frac{1}{y}$) for y $\neq$ 0, then we have the inequalities

-|y| $\leq$ G(y) $\leq$ |y| ;for all y $\neq$ 0

Followed by well known inequality |Sin y| $\leq$ 1 $\forall$ y $\in$ R

=> Lim$_{y \rightarrow 0}$ (-|y| ) = 0 and Lim$_{y \rightarrow 0}$ (|y| ) = 0

By Squeeze theorem,   Lim$_{y \rightarrow 0}$ y sin ($\frac{1}{y}$) = 0

Example 2: Let g(y) be a function s.t. |g(y)| $\leq$ L for any y $\neq$ 0. Prove that $lim_{y \rightarrow 0}$ yg(y) = 0

Solution: Since |g(y)| $\leq$ L for any y $\neq$ 0.

or |y g(y)| $\leq$ L|y|

=> -L|y| $\leq$ y g(y)| $\leq$ L|y|

For any non zero y,

$lim_{y \rightarrow 0}$ (-L|y|) = 0 = $lim_{y \rightarrow 0}$ (L|y|)

So by squeeze theorem, $lim_{y \rightarrow 0}$ yg(y) = 0

## Squeeze Theorem for Sequences

Squeeze theorem used to find the convergance and divergence of the sequences. If $lim_{n \rightarrow \infty}$ $a_n$ = $lim_{n \rightarrow \infty}$ $b_n$  = L and there exists an integer N s.t. $a_n$ $\leq$ $c_n$ $\leq$ $b_n$ for all n > N, then $\lim_{n \rightarrow \infty}$ = L.

Let's understand this concept in a better with the help of an example:

Example: Determine if the below sequence if converges or diverges.
{$b_n$ } = {$\frac{sin n}{n}$}

Solution: We know that, for  all n, 1 $\leq$ sin n $\leq$ 1

Therefore, -$\frac{1}{n}$ $\leq$ $\frac{sin n}{n}$ $\leq$ $\frac{1}{n}$

Choose $a_n$ = -$\frac{1}{n}$ and $c_n$ = $\frac{1}{n}$

Also $a_n$ $\leq$ $c_n$ $\leq$ $b_n$ for all n

=> $lim_{n \rightarrow \infty}$$a_n = 0 = lim_{n \rightarrow \infty}$$b_n$

So by sequeeze theorem $lim_{n \rightarrow \infty}$$c_n$ = 0

This implies, the sequence {$b_n$ } is converges to zero.